Question

In a triangle $ABC,<C={90}^{\circ }.$ Find the value of ${\sin }^{2}A+{\sin }^{2}B$

Answer 1

In $△ABC$ $\angle C=90$

To find ${\sin }^{2}A+{\sin }^{2}B$

$\Rightarrow \sin A=\frac{P}{H}=\frac{BC}{AB}$ $⟶\left(1\right)$

$\Rightarrow \sin B=\frac{AC}{AB}$

$\Rightarrow {\sin }^{2}A+{\sin }^{2}B={\left(\frac{BC}{AB}\right)}^2+{\left(\frac{AC}{AB}\right)}^2$

$=\frac{B{C}^2+A{C}^2}{A{B}^2}$

since $\angle C=90$

$\therefore$ By using Pythagoras theorem $

$\Rightarrow A{B}^2=B{C}^2+A{C}^2$

$\Rightarrow {\sin }^{2}A+{\sin }^{2}B=\frac{A{B}^2}{A{B}^2}=1$

$\Rightarrow {\sin }^{2}A+{\sin }^{2}B=1$

Hence, the answer is ${\sin }^{2}A+{\sin }^{2}B=1.$